Question

Find the value of x for the expression, \[\log \left( x+1 \right)+\log \left( x-1 \right)=\log 11+2\log 3\]

Answer 1

Hint: Use the properties of logarithm, \[b\log a=\log {{a}^{b}}\] and \[\log mn=\log m+\log n\] to simplify the given expression. Now, solve it further and get the value of x. At last, check all the logarithmic terms by putting the value of x in the given expression. Finally, ignore the value of x for which the term inside the logarithm is negative.

Complete step-by-step answer:
According to the question, we are given an expression in which there is one unknown variable \[x\] and we are asked to find the value of \[x\] .
The given expression is \[\log \left( x+1 \right)+\log \left( x-1 \right)=\log 11+2\log 3\] ………………………………………(1)
We can observe that the above equation is somewhat complex, so we need to transform it into a simpler form.
 We know the property of logarithm, \[b\log a=\log {{a}^{b}}\] ……………………………………(2)
Now, on putting \[b=2\] and \[a=3\] in equation (2), we get
\[\Rightarrow 2\log 3=\log {{3}^{2}}\]
\[\Rightarrow 2\log 3=\log 9\] …………………………………………………(3)
From equation (1) and equation (3), we get
\[\log \left( x+1 \right)+\log \left( x-1 \right)=\log 11+\log 9\] ……………………………………………(4)
We also know the property of logarithm, \[\log mn=\log m+\log n\] …………………………………………..(5)
Replacing \[m\] by \[\left( x+1 \right)\] and \[n\] by \[\left( x-1 \right)\] in equation (5), we get
 \[\log \left( x+1 \right)\left( x-1 \right)=\log \left( x+1 \right)+\log \left( x-1 \right)\] …………………………………………….(6)
 Similarly, replacing \[m\] by 11 and \[n\] by 9 in equation (5), we get
 \[\log \left( 11 \right)\left( 9 \right)=\log \left( 11 \right)+\log \left( 9 \right)\] …………………………………………….(7)
Now, from equation (4), equation (6) and equation (7), we get
 \[\begin{align}
  & \Rightarrow \log \left( x+1 \right)\left( x-1 \right)=\log 11\times 9 \\
 & \Rightarrow \log \left( {{x}^{2}}-x+x-1 \right)=\log 99 \\
\end{align}\]
\[\Rightarrow \log \left( {{x}^{2}}-1 \right)=\log 99\] ……………………………………………..(8)
On applying antilog in LHS and RHS of the above equation, we get
 \[\Rightarrow \left( {{x}^{2}}-1 \right)=99\]
\[\Rightarrow {{x}^{2}}=100\]
So, \[x=\pm 10\] …………………………………………….(9)
But, \[x=-10\] is not a possible value of x because putting \[x=-10\] will make the term inside the logarithm \[\log \left( x-1 \right)\] negative and the term inside the logarithm cannot be negative.
Therefore, \[x=-10\] is not possible.
Hence, the possible value of x is equal to 10, \[x=10\] .

Note: In this question, one might include \[x=-10\] as the possible value of x. This is wrong. So, whenever a question appears where there is logarithmic related terms. Always check the value for which the logarithmic terms is positive or negative. At last, ignore all those values for which the logarithmic term is negative. A common silly mistake can be made by the students in the logarithmic property used, instead of multiplying the terms, they may divide the terms.